Problem: $\overline{AC}$ is $9$ units long $\overline{BC}$ is $5$ units long $\overline{AB}$ is $\sqrt{106}$ units long What is $\cos(\angle BAC)$ ? $A$ $C$ $B$ $9$ $5$ $\sqrt{106}$
Answer: SOH CAH TOA os = djacent over ypotenuse adjacent $= \overline{AC} = 9$ hypotenuse $= \overline{AB} = \sqrt{106}$ $\cos(\angle BAC )=\frac{9}{\sqrt{106}}$ $=\dfrac{9\sqrt{106} }{106}$